Kernel of an algebra map and module of Kahler Differentials

Question

Let $A$ be a $k$-algebra, $f:A\rightarrow k$ an algebra map with kernel $I$.

I'd like to prove that $\Omega_A\otimes_f k $ is canonically isomorphic to $I/I^2$.

This is from W.C.Waterhouse Intro to Aff. Group Schemes pag 85 (d). I can get that given a $k$-linear derivation $D$ from a $A$ to some $k$-algebra $N$ upon which $A$ acts by $f$, $D$ must vanish on $I^2$ and so must factorize along the projection $p:A\rightarrow A/I^2$, but then what?

Related question: Kahler differentials of a Hopf Algebra

Answer 1

I give here the sketch of my proof for the result:

We have from the exact sequence associated to the surjective map :→ the surjective morphism $\delta:I/I^2 \rightarrow Ω_⊗_ $ , which induces the injective morphism Der$_k(A,N) \rightarrow$ Hom$(I/I^2, N)$ for every $k$-algebra $N$ made into an $A$-algebra by $f$. If the induced morphism is also surjective for every $N$, our $\delta $ will be an isomorphism as we want.

So we just have to prove that Der$_k(A,N) \rightarrow$ Hom$(I/I^2, N)$ is surjective for every $N$:

For every element in , you can define an element $_=−\rho()$ (where $\rho$ is just the k-algebra structure of A), and $i_a$ belongs to because $f\rho=id$; so you get a linear map $ \rightarrow/^2 $. By composing with this map you get a derivation $A\rightarrow I/I^2 \rightarrow N $ for every $k$-linear morphism in $Hom_k(I/I^2, N)$ , as we wanted

Answer 2

Let $s:k\to A$ be the structure morphism of $A$ as $k$-algebra, I mean $s(\lambda)=\lambda\cdot1_A$ for any $\lambda\in K$, where $1_A$ is the (multiplicative) unit of $A$. By hypothesis, $f$ is a morphism of $k$-algebras; because \begin{equation} \forall\lambda\in k,\,f(s(\lambda))=f(\lambda\cdot1_A)=\lambda\cdot f(1_A)=\lambda\cdot1=\lambda \end{equation} that is $s$ is section of $f$, by Stacks Project TAG 02HP one has: \begin{equation} \Omega^1_{A/k}\otimes_fk\cong I_{\displaystyle/I^2} \end{equation} where $I=\ker f$.